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					                                                                         n° 162
                                                                         electrodynamic
                                                                         forces on
                                                                         busbars in
                                                                         LV systems
Jean-Pierre Thierry                   Christophe Kilindjian

Engineering graduate from the         After graduating as an engineer
CESI (Centre d'Etudes                 from the Ecole Supérieure
Supérieures Industrielles) and        d'Energie et des Matériaux of
from the CNAM (Conservatoire          Orléans in 1986, he then joined
National des Arts et Métiers),        Merlin Gerin in this same year
he was initially employed in the      as part of the Technical Section
Iron and Steel industry (roll mill    in the Low Voltage
automation and fluid                  Switchboards unit.
monitoring).                          Responsible for basic studies,
Following several years               he specialises in problems of
dedicated to preparing and            heat exchanges and
developing mechanical                 electrodynamic withstand in LV
vibration testing means, he           equipment.
joined Telemecanique in 1969.
He then held in turn the
positions of engineering and
design department manager
and technical Project manager
for new products. He is
currently in charge of
developing prefabricated
electric ducts.




E/CT 162 first issued, october 1996
Cahier Technique Merlin Gerin n° 162 / p.2
electrodynamic forces on busbars in                                              The importance attached to the
                                                                                 concept of industrial dependability
LV systems                                                                       (safety of persons and equipment,
                                                                                 availability of electrical power,
                                                                                 reliability and maintenability of
                                                                                 products) increasingly affects the
                                                                                 design of the electrical devices used
                                                                                 in industry (process...) and tertiary
                                                                                 (hospitals). Their operating
                                                                                 dependability thus contributes, often
                                                                                 to a large context, to the
                                                                                 dependability of the installation as a
contents                                                                         whole. This is the case of low voltage
                                                                                 (LV) switchboards and of
                                                                                 prefabricated transformer-
1. Introduction                                                          p. 4    switchboard connections.
2. Electrodynamic forces between two   Preliminary remarks               p. 4    This quest for dependability requires
conductors: origin and calculations                                              studies in order to master, from the
                                       Origin and calculation methods    p. 4
                                                                                 design stage, the behaviour of their
                                       Calculation for two parallel      p. 6    components in the light of their
                                       filiform conductors of infinite           environment and of possible operating
                                       length                                    stresses. One of these studies has
                                       Influence of conductor shape      p. 6    already been dealt with in a «Cahier
                                       Conductors of reduced length      p. 7    Technique» (thermal behaviour of
                                                                                 LV electric switchboards). Withstand of
                                       Non-rectilinear conductors        p. 7    electrodynamic forces is now the
                                       Calculation in the case of        p. 7    subject of a second study.
                                       complex configurations                    Designers will find in this «Cahier
3. Electrodynamic forces in a          Reminder on short-circuit         p. 9    Technique» the calculations laid down
three-phase busbar on a two or         current making                            to allow for these forces and in
three-phase fault                                                                particular to determine LV busbar
                                       Maximum force on a                p. 10
                                                                                 requirements (prefabricated in ducts for
                                       three-phase busbar
                                                                                 electrical power distribution, and in
                                       Resonance phenomena               p. 11   switchboards).
4. Application to LV three-phase       Case of busbars in                p. 12   However calculation alone is not
busbars                                LV switchboards                           sufficient and results need to be
                                       Case of prefabricated ducts       p. 16   validated by a real-life tests. We thus
                                       of the Canalis and Victa Dis type         briefly describe the standardised tests.

5. Conclusion                                                            p. 20
6. Bibliography                                                          p. 20




                                                                                 Cahier Technique Merlin Gerin n° 162 / p.3
1. introduction


The problem of withstanding                   between the various conductors of a          geometry of the conductors and
electrodynamic forces arises on the           LV installation (solid conductors of the     associated structures.
LV power circuits of the installation.        bar, cable type...) generate considerable    However a few approximations yield in
Although mainly dependent on the              forces (several thousands of daNm).          most cases valid results on the basis of
strength of the fault current, it also        These forces thus need to be                 simple formulae.
depends on the shape of the                   determined in order to mechanically          After a brief reminder of calculation of
conductors, their mutual setout and           size both the conductors and the             electrodynamic forces in simple
securing method. Although this problem        structures supporting them so that they      geometries, this Cahier will deal with
can be solved by calculation, only            can withstand these forces whatever          busbars in switchboards and
validation by a real-life tests enables       protective devices are placed upstream       prefabricated ducts on the basis of
provision of a document acknowledging         and downstream (standards stipulate          these formulae.
conformity with standard and/or               electrodynamic withstand tests of one
customer requirements.                        second).
The very high current strengths that          The exact calculation of electrodynamic
may occur during a short-circuit              forces is often complex in view of the




2. electrodynamic forces between two conductors:
origin and calculations


The problem of conductor withstand to         However the root mean square values          whether between two current
electrodynamic stresses is certainly not      Irms are used in most cases; in this         elements or between a magnetic field
new as is shown by the number of              case Irms must be multipled by a             and an electric current (work
publications which have treated this          coefficient defined in chapter 3.            conducted by Oersted, Ampère...)
issue. However this problem is still of       c forces are expressed in absolute           resulted in the construction of a
interest to designers as a result of the      value without specifying their direction     theoretical framework integrating
application of modern numerical               depending on field and current               these dynamic phenomena between
methods which provide a solution for          direction.                                   conductors through which electric
complex conductor configurations. This        In most cases they are forces per unit       current flows.
accounts for the summary presented in         of length.                                   The direction of the electrodynamic
this chapter.                                 c conductors are made of non-                forces is known (repulsion if the
                                              magnetic material and are sufficiently       currents in the conductors flow in
                                              distant from all magnetic elements
preliminary remarks                                                                        opposite directions, otherwise
                                              likely to alter distribution of the          attraction) and their values are
Application of the formulae call for          magnetic field that they create.
compliance with the following points:                                                      obtained by applying the laws of
                                              c skin effect and proximity phenomena
c all the formulae involve the product of                                                  magnetism.
                                              which can considerably alter current
the current strengths, I1.I2, flowing in                                                   There are in fact two main methods
                                              distribution in the cross-section of solid
each conductor and inter-reacting. If                                                      for calculating electrodynamic forces.
                                              conductors are ignored.
their values are identical, this product is                                                The first method consists of
replaced by the term I2.                                                                   calculating the magnetic field created
c the current strengths appearing in the
                                              origin and calculation                       by an electric current at a point in
formulae correspond to the peak value         methods                                      space, then deducing from it the
of the currents conveyed in each              The highlighting and understanding a         resulting force exerted on a conductor
conductor.                                    century ago of mutual influences,            placed at this point and through which




Cahier Technique Merlin Gerin n° 162 / p.4
an electric current flows (possibly             or Ampère's theorem:                                  The second method is based on
different from the first one).                        →                                               calculating the potential energy
To calculate the field it uses (see box,
                                                ∫B   d l = µ0 I ,
                                                                                                      variation of a circuit and uses Maxwell's
                                                c
fig. 1) either Biot and Savart's law:                                                                 theorem :
                                                and to calculate electrodynamic force, it
                     →       →                                                                                     δΦ
      →  µ    dl∧ u                             uses Laplace's law:                                   (4) Fx = i
                                                                                                                   δx
(1) d B = 0 i       ,                                 →       →     →
         4π    r2                               (3) d f = i d l ∧ B .                                 (see box, figure 1).




    Biot and Savart's law                                                  Ampère's theorem
    Each element of a circuit through which a current i
                        →                                                  Deduced from Biot and Savart's formula, it is
    flows, of a length d l , produces at a point M a field
       →
                                                                           expressed as follows:
     d B such that:                                                        Let I be the current strength flowing through a
                         →   →                                             conductor crossing any surface of contour C.
      →    µ0 d l ∧ u                                                      Circulation of the magnetic field along C is given by the
    dB =      i       .
           4π   r2                                                         equation:
    This field is:                                                                →
    c→ perpendicular to the plane defined by the element                   ∫ B d l = µ0 I .
                                                                            c
     d l containing point P and point M,
    c oriented to the left of an observer placed on the                    Laplace's law
    element, with the current flowing from his feet to his                 When a circuit through which a current of strength i
    head and his gaze directed to point M (Ampère's                                                                →
                                                                           flows, is placed in a magnetic field B , each element
                                                                              →
    theorem)
                                                                            d l of the circuit is subjected to a force equal to:
                     →           →                        →                     →        →    →
    c modulus d B where u is the directing vector of PM .                  d f =i d l∧B
                                                                                  →
                                                                           When B has an electric circuit as its origin, the law
                                                                           applied to each one expresses the force exerted
                                                                           between them:
                                                                                →         →       →        →   →
                     i                                                      d f = i1 d l ∧ B 2 = i2 d l ∧ B1 .

                                                                           Maxwell's theorem
                                                      →                    The work of the electromagnetic forces exerted during
                                                     dB                    displacement of an undeformable conductor through
                 →                                                         which an invariable current flows, placed in a magnetic
                dl               →                                         field, has the following expression:
                             θ   u
                                                                           w = i Φ or Φ is the flow of the magnetic field swept
                                            M
                     P               r                                     during the displacement.
                                                                           Used in the form of elementary work, it easily obtains
                                                                                                                           →
                                                                           the components Fx, Fy and Fz of the resultant F of the
                                                                           electromagnetic forces:
                                                                            dw = i dφ
                                                                                         →    →
                                                                                    = ∫d f d l
                                                                                     →    →
                                                                                    = F d l hence
                                                                                         δΦ
                                                                           Fx = i
                                                                                         δx
                                                                           and likewise for Fy and Fz.



fig. 1: reminder of physical laws.




                                                                                                      Cahier Technique Merlin Gerin n° 162 / p.5
According to the geometry of the               always valid. In this case the influence               F/ l in N/m,
conductor system considered and to             of conductor shape may be determined                   I1 and I2 in A,
the calculation difficulty, one of the         by considering the conductor cross-                    d in m.
three approaches (1)+(3), (2)+(3) or (4),      section as a superimposition of inter-                 Examples of forces withstood by two
can be used.                                   acting current lines. This approach was                parallel bars on a short-circuit are given
However the results obtained may differ        made by Dwight for a conductor with a                  in the table in figure 3.
slightly according to the approach used        rectangular cross-section.
                                                                                                      Although the same approach can be
since the assumptions on which these           The resulting corrective factor,
                                                                                                      followed for all conductor shapes,
laws are established are not the same.         conventionally denoted k, can be
                                                                                                      calculations quickly become tiresome.
                                               determined by calculation. However as
                                               the expression of k is relatively complex,             In the above equation the term (k/d) is
calculation for two parallel                                                                          often replaced by 1/D, where D stands
                                               its value is determined in most cases on
stranded conductors of                         the S-shaped curves as in figure 2.                    for the distance between the
infinite length                                                                                       conductors corrected to allow for the
                                               The equation then has the form:
                                                                                                      influence of their shape.
For simple geometries such as filiform
                                               F/ l = 2 10-7 I1 I2 (k/d)                              These coefficients are also useful in the
rectilinear conductors, application of
Biot and Savart's and of Laplace's law         where :                                                case of a set of three-phase conductors
results in the classical formula for
electrodynamic force between two
current lines;
 F/ l = 2 10-7 I1 I2/d
                                                 k
where :
                                               1.4
F/ l in N/m,                                                                                                                a
I1 and I2 in A,
d in m,                                        1.3
                                                                  b                                                         b
(The coefficient 2 x 10-7 results from                 0.01...0.2 a
the ratio µ0/4 π).                             1.2
As this formula acts as a basis                                                                                    d
throughout this study, we must specify
                                               1.1
the assumptions for which this
                                                     0.5                                                  a
expression is valid.                                                                                                            b
c the conductors are reduced to a                1
current line. Their cross-section is thus             1
reduced to a point. In practice this
                                               0.9
condition is considered acceptable for
conductors of all cross-sections if the
                                                           2




distance between the two conductors is         0.8
considerably larger than the largest
                                                                    5




transverse dimension of the conductors
                                                                           10




                                               0.7
(e.g. ten times).
                                                                                  15
                                                                                20




c the conductors are considered to be
                                                                                                 30




                                               0.6
                                                                                                40




rectilinear and infinitely long. In practice
                                                                                              50
                                                                                             60




this condition may be considered
                                                                                           80
                                                                                          0
                                                                                        10




satisfactory if they are at least 15 to 20     0.5
times longer than the distance between                                                                                 b
                                                                                                                       a
them.
Whenever one of these assumptions is           0.4
not valid, a corrective factor must be
applied.                                       0.3


influence of conductor                         0.2                                                                                            d
                                                                                                                                              a
shape                                                          2    3    4 5 6      8 10         20           40       60 80 100    200 300

This formula of F/ l only applies to
current lines. However for solid
                                               Fig. 2: variation of k as a function of ratios b/a and d/a (Dwight's chart).
conductors this assumption is not




Cahier Technique Merlin Gerin n° 162 / p.6
containing several conductors per                         This formula can be used only for
phase. This case is dealt with in                         values of a and b
chapter 3.
                                                                             a
                                                          such that: 1 <       < 10                                            I
                                                                             b
conductors of reduced
length                                                    calculation in the case of
c conductors of identical length                          complex configurations                                                                                              d
When conductors have the same lengt
l , 15 to 20 times smaller than their                     The busbar configurations considered
centre distance d, the resulting force is:                up to now in this study were mainly
                                                          mono-dimensional, or sometimes two-
                      l                  d
                                                          dimensional in the case of conductors                           c1                     l           c2
                                   d2
F = 2 10 −7 I 2              1+         −               forming an angle. In these cases, the
                      D           l2     l
                                                        methods used to calculate
c conductors of unequal length (see                       electrodynamic forces result in
                                                                                                                  fig. 4: drawing showing two conductors of
fig. 4)                                                   relatively simple formula.
                                                                                                                  unequal length.
In this case the resulting force is:                      However conductors can be arranged
                                                          in many different ways or be associated
                      l
F = 2 10 −7 I 2
                      D
                          [C1+ C2]                        with a «disturbing» environment, such
                                                          that the above formulae cease to apply.
where                                                                                                                                        2
                                                                                                                                    d2                       c2          d2
                                                                                                                        C =  1+  + 2 −
                                                          Such arrangements are referred to as                                  c
                                                                                                                                                                     +
                                                          «complex configurations».                                          l    l                        l   2
                                                                                                                                                                         l2
                  2
     c      d           2        c12           d   2
C1=  1+ 1  + 2 −                           +
        l   l                    l 2
                                                 l2                                                               C                                                      c
                                                                                                                                                                         l
and
                                                                                                                   1                                         ∞
                  2                                                                      d                        0.9                                    2
      c      d2                   c22              d2
C2 =  1+ 2  + 2 −                              +                                                                0.8                                1
         l   l                        l2           l2
                                                                                                                  0.7                        0.5
The values of C1 and C2 can be read                                                                                                    0.2
                                                                                 l                                0.6
                                                                                                                                   0
on the chart in figure 5.                                 F                                                   F   0.5
If the conductors do not face each other                                                                          0.4
over the entire length, with one passing                                                                          0.3
the other, the formula applies with c1 or                     b
                                                                                                  I               0.2                                                             d
c2 negative.
                                                                                                                        0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2                       l
NB
                                                                   a
If c / l = 0, the equation is F in the
above paragraph. The value of the
                                                                                                                  fig. 5: calculation and variation of C as a
expression between square brackets is
                                                                                                                  function of the ratios c / l and d / l .
given directly by reading the relevant
curve on the chart in figure 5.                           characteristics                    forces

                                                          a       b    d      l      k       I        F
non-rectilinear conductors
This is, for example, the case of                         mm mm        mm    m               kA       daN/m                                                                       F
conductors with a bend (see fig. 6). The                                                                                                             b
                                                          5       80   100   1       0.91 35          224
branches may inter-act with one another
when a strong current passes through                      5       80   100   1       0.91 80          1170         I                                                 α
them.                                                                                                                      a
The conductor b may pivot around point
O of the fixed conductor a. Force F has                                                                                                      O
                                                          fig. 3: characteristics required to calculate
the following value:
                                                          the forces F between two conductors of the
                                                          same length.
                  a   a    b 2  1- cosα                 Examples of forces withstood by two parallel            fig. 6: drawing showing two part of
F = 2 10-7 I 2      l  + 1+ 2 
                  b   b    a  sinα                      bars during a short-circuit.                            conductors (a and b) with a bend.
                               




                                                                                                                  Cahier Technique Merlin Gerin n° 162 / p.7
Three types of problems may then             numerically solve the problems               c choice of meshing parameters and
arise, either separately or combined:        described by differential equations. In      meshing of the calculation scope with
c the conductors facing one another          particular the finite elements method,       the type of elements chosen; at this
are not all in the same plane: the           initially developed for mechanical           stage, the system studied is merely a
problem is three-dimensional;                problems, has been extended to a wide        set of nodes;
c the conductors are close to metal          range of sectors and notably that of         c definition of boundary conditions to
frames which may alter the distribution      electromagnetism.                            solve the equations;
of the magnetic field surrounding them;      In short, to define the calculation scope,   c carrying out the calculation;
c the conductors are arranged so that it     this method consists of breaking down        c using the results.
may be necessary to allow for skin           the system studied into a certain            A wide range of calculation software is
effect and proximity phenomena which         number of elements constituted and           available, differing by the categories of
may considerably alter current               connected with one another by points         problems they can solve and the
distribution in the cross-section of solid   known as nodes. The quantities which         reliability of the results that they yield.
conductors.                                  are of interest to us (magnetic field,       For example Merlin Gerin has chosen
Calculation of electrodynamic forces for     stresses) are determined numerically at      the ANSYS software and
the three types of problems mentioned        each node by salving the relevant            Telemecanique Flux 2D since:
above uses the general approach              equations (Maxwell and elasticity).          c they enable very different problems to
described in the paragraph on «origin        Consequently, the value of each              be dealt with (thermal, mechanical,
and calculation method», namely              quantity studied is not known exactly at     electromagnetism...),
calculation first of the value and           all points of the system but only at node    c they are open-ended; thus their latest
distribution of the magnetic field at each   level. Hence the importance of ensuring      versions enable different problems to
point in the system, then of the stresses    a good correspondence between these          be paired (magnetic and mechanical or
in the conductors. The problem is thus       nodes and the real system, and of            mechanical and thermal...).
divided into two to yield a magnetic and     having a sound meshing. In practice,         It is true that these methods may seem
a mechanical problem.                        calculation using this method is made        cumbersome and call for considerable
The basic physical laws used are             up of the following stages:                  investment. However with thorough
therefore the same. However the              c choice of the analysis type                mastery of the problems relating to
difficulty, compared with the simple         (e.g. magnetism...);                         modelling techniques, they allow rapid
cases, lies in performing the                c choice of type of elements to              evaluation of the behaviour of a system
calculations, as the three-dimensional       describe the system;                         or of one of its parts other than by tests.
aspect requires a numerical approach.        c definition of the system geometry and      This is especially appreciable during the
Numerous methods have been                   of the calculation scope using key           design and development phases when
developed in recent years to                 points;                                      you consider the cost of a test campaign.




Cahier Technique Merlin Gerin n° 162 / p.8
3. electrodynamic forces in a three-phase busbar
on a two or three-phase fault


Consideration of three-phase busbar                  of these forces and the conductor with      isolated two-phase faults which have
peculiarities when designing busbars                 the highest mechanical stress.              the advantage, in steady state, of
for LV switchboards and prefabricated                As the electrodynamic forces of the         behaving like one or two independent
ducts, and of the peculiarities relating to          current are proportional to the square of   single-phase networks.
the establishment and type of fault, is              its maximum amplitude, the short-circuit    Let us consider a fault occurring on the
achieved by integrating factors into the             currents need to be studied.                single-phase diagram in figure 8 in
formule presented in chapter 2.
                                                                                                 which R and L ω are network
These peculiarities are:
c relative layout of phases (conductors
                                                     reminder on short-circuit                   impedance elements. If we set as the
                                                                                                 origin of time, the moment when the
in ribbon, staggered...),                            current making                              short-circuit occurs, the e.m.f. (e) of the
c phase shift of currents in each phase              The aim of this paragraph is to review      generator has the value:
with respect to one another,                         and specify:
c type of short-circuit (two or three-               c the various short-circuit types that      e = 2 E sin( ω t + α )
phase),                                              can arise in a three-phase system,
c short-circuit making characteristics               c the notions of symmetrical and            where α is the energising angle (see
(symmetrical or asymmetrical state),                 asymmetrical state,                         fig. 9) corresponding to the offset in
c the peak current value,                            c the procedure to follow to determine      time between a zero of the e.m.f. and
c the alternating aspect of currents,                the expression of short-circuit currents    the moment when the short-circuit was
hence the vibrating aspect of the                    and the parameters on which they            made.
phenomena they generate.                             depend.
In the remainder of this section, the                                                            Ohm's law applied to the circuit yields:
                                                     The short-circuit types
study will consider only busbars in                                                                          di
ribbon, where phases 1,2, 3 are set out              There are four types on a three-phase       e = R i+L
                                                     network. These types are shown in                       dt
in the same plane and with the same
distance between phases.                             figure 7.                                   If the current is nil before the short-
                                                     Expression of short-circuit currents        circuit is made, the solution for this
The aim is to determine, by analysing
                                                     in the case of a three-phase fault          equation is:
the change in electrodynamic forces as
a function of time and the various
parameters above, the maximum value
                                                     We shall now concentrate only on
                                                     symmetrical three-phase faults and                          [
                                                                                                 i( t ) = 2 I sin( ω t + α - ϕ ) + sin( ϕ - α ) e -t / τ   ]

a)         L3                                        c)         L3                                           R             x
           L2                                                   L2
           L1                                                   L1                                                                          A

                                                                                                                     Zcc                           Z1
                              I"
                               k
                                                                     I"
                                                                      k             I"
                                                                                     k
                                                                          I"
                                                                           k
                                                                                                                               B
                                                                                                 fig. 8: equivalent single-phase diagram on a
b)         L3                                        d)         L3                               three-phase fault (see IEC 909).
           L2                                                   L2
           L1                                                   L1
                                                                                                 u                               u = f(t)
                              I"
                               k
                                                                               I"
                                                                                k



                                                                                                                                                    ωt
                                                                                                                      α
a) symmetrical three-phase short-circuit.            c) short-circuit between phases, with
b) short-circuit between phases, isolated or         earthing.
two-phase.                                           d) phase-earth short-circuit
fig. 7: the various short-circuits and their currents. The direction of the arrows showing the   fig. 9: representation of α known as the
currents are random (see IEC 909).                                                               energising angle.




                                                                                                 Cahier Technique Merlin Gerin n° 162 / p.9
where :                                            by side. Thus each conductor                  F/ l = 2 10-7 I1 I2/d
              Lω                                   undergoes at a time t a force which
ϕ = arctg            (impedance angle)             results from the algebraic addition of its   the additional corrective factor which,
               R                                                                                according to the position of the
                                                   interactions with the two other
     L                                                                                          conductor under consideration, equals
τ=                                                 conductors. These conductors can
     R                                             have only two situations, external or        0.808 or 0.866. The maximum force is
              E                                                                                 thus generated on the central
I=                                                 central:
         R +L2 ω 2
          2
                                                   c external position, for example             conductor.
                                                   phase 1:                                     c in practice, the coefficient k takes the
All the factors representing the current                                                        circuit characteristics (R and L) into
                                                   F1(t) = F2→1(t) + F3→1(t)
variation as a function of time are then                                                        consideration: its value is between 1
grouped in the following equation:                 F1(t) = cF [ i1(t) i2(t) + i1(t) i3(t)/2]    and 2 (see fig. 10).
                                                   cF is a function of distance between
     [                                         ]
                                                                                                Case of a two-phase short-circuit
κ = sin(ω t + α - ϕ ) + sin( ϕ - α ) e-t / τ       bars and bars shape.                         In this case i1 = - i2 and, using the
                                                   c central position, for example phase 2:     above formulae, we can show that the
The term κ can also be calculated                                                               maximum electrodynamic forces are
                                                   F2(t) = F1→2(t) - F3→2(t)
using the approximate formula defined                                                           reached when α = 0 (asymmetrical
by IEC 909:                                        F2 = cF [i1(t) i2(t) - i2(t) i3(t)]
                                                                                                state).
                          3R                       However, as seen in the above                F2max ,2ph
                      -
κ = 1.02 + 0.98 e         Lω                       paragraph, there are many cases to be
                                                   considered for current expression            = 2 10-7 1 ( 2 Irms,2ph κ)2 1/d
The difference with the exact value is             according to the value of α, ϕ, and of
less than 0.6%.                                                                                 Remarks
                                                   the type of short-circuit.                   The maximum force is not shown in
Analysis of this function enables                  In actual fact, only the value of the        two-phase, as it is often thought, but in
definition of the symmetrical and                  maximum forces is required to size the       three-phase.
asymmetrical states of a fault
                                                   busbars: this value is the highest           In actual fact:
(cf. «Cahier Technique» n° 158).
                                                   current occurring when α = 0.
In the case of a three-phase system,                                                            F2max,3ph       0.866 I 2 rms,3ph
the current in each phase takes the                NB:                                                      =
                                                   Fa—>b = action (force) of the                F2max,2ph          I 2 rms,2ph
form:
                                                   conductor(s) of phase a on the
                          [
                                                                                                however in the three-phase distribution
i1( t ) = 2 Irms,3ph sin(ω t + α - ϕ )             conductor(s) of phase b.                     state:
          +sin(ϕ - α ) e-t/ τ   ]                  Case of a three-phase short-circuit
                                                                                                Irms,2ph =
                                                                                                            3
                                                                                                              I
which can also be written as:
                                                   The effects on the conductors take the                  2 rms,3ph
                                                   form:
                                                                                                which yields the ratio:
i1( t ) = 2 Irms,3ph κ                             F1 = 0.87 [i1(t) i2(t) + i1(t) i3(t)/2]
where Irms,3ph stands for the                      F2 = 0.87 [i1(t) i2(t) - i2(t) i3(t)]        F2max,3ph
                                                                                                             ≈ 1.15
symmetrical root mean square current               The maximum force on conductors over         F2max,2ph
in the three phases in steady state.               time is determined by the time values
In view of their relative phase shift:             which cancel the derivatives of these
c i2 same as i1 by replacing α by                  expressions with respect to time:
α + 2π/3                                           dF1/dt = 0 et dF2/dt = 0.
c i3 same as i1 by replacing α by                  Hence, after a few calculations, where         k
α - 2π/3.                                          Imax,3ph = 2 Irms,3ph κ                      2.0
Finally, the electrodynamic forces thus            the two equations:
depend on :                                                                                     1.8
                                                   c F1max ,3ph =
c the initial instant of the short-circuit
(via the value of α) ;                             2 10- 7 0.808 ( 2 Irms,3ph κ)2 1/d           1.6

c the characteristics of the circuit (via          (case of one of the conductors external      1.4
the value of ϕ) ;                                  to the three-phase busbar)
c the phase shift between the phases               c F2max ,3ph =                               1.2
(2π/3).                                            2 10-7 0.866 ( 2 Irms,3ph κ)2 1/d            1.0
                                                                                                                                             R/X
                                                                                                      0   0.2   0.4    0.6   0.8    1.0   1.2
                                                   (case of one of the conductors external
maximum force on a three-                          to the three-phase busbar)
phase busbar                                       Note:
                                                                                                fig. 10: variation of factor k as a function of
A three-phase busbar normally                      c compared with the reference formula
                                                                                                the ratio R/X.
contains three conductors placed side              reviewed in chapter 2




Cahier Technique Merlin Gerin n° 162 / p.10
Test organisations often demand two                         δy 2    δy     δy 4                      expression of conductor natural
                                               F( t ) = M        +λ    +E J 4
and three-phase tests with currents of                      δt 2    δt     δx                        resonance frequencies:
identical value. These test conditions                                                                          2
                                               where:                                                        Sk E J
do not correspond to real distribution                                                                ω ok = 2
                                               M = mass of the conductor per unit of                          p    M
characteristics and result in two-phase
                                               length,                                               where
forces which are greater than three-
                                               J = moment of inertia of the cross-                   Sk = coefficient function of the securing
phase forces.
                                               section perpendicular to the conductor                methods, for example for a bar flush
                                               axis,                                                 mounted at its ends:
resonance phenomena                            E = modulus of elasticity,
                                                                                                     Sk = (4 k - 1) π/2 ;
Forces appearing on a short-circuit do         λ = damping coefficient,
                                               y = distance from a point of the                      k = rank of resonance frequency;
not form a static phenomenon, but are
vibrating quantities of a frequency twice      conductor with respect to its position of             p = distance between the supports.
that of the network or of its multiples.       equilibrium or deflection,                            In practice, we observe that natural
Conductors which have a certain                x = distance from a point of the                      conductor frequencies, for a specific
elasticity can then start to vibrate. If the   conductor with respect to a fixed                     cross-section, depend on the
vibration frequency corresponds to a           bearing,                                              longitudinal distance between supports.
natural frequency for all conductors,          t = time.                                             The calculation therefore aims at
resonance phenomena may occur. In              where F(t) = Fo sin (2 ω t)                           examining whether the stress factor,
this case the resulting stresses in the        where:                                                resulting from the selected distance
conductors may be far greater than             Fo = amplitude of the force,                          between the supports, is acceptable for
                                                                                                     the natural frequency of the conductor
those created by the forces due to the         ω = network pulsation (ω = 2 π f).
peak current value. It is thus necessary                                                             or all the conductors, resulting in a
                                               The solutions take the form:                          coefficient R, homogenenous with a
to determine the ratio between the real
                                               y = cste Fk(t) Gk(x)                                  length:
and static forces undergone by the
                                               where the functions Fk(t) and Gk(x)
conductor. This ratio conventionally
                                               depend on time and on the space                               EJ
denoted Vσ is known as the stress                                                                    R=4           103
factor. In addition to the mechanical
                                               variable respectively, as well as on:                        M ω2
                                               c the securing methods,                               The graph in figure 12 shows the stress
characteristics of the conductors, we
                                               c the electrodynamic force relating to                factor Vσ to be anticipated as a function
must allow for the way in which they
are secured in the device housing them         the short-circuit state (symmetrical or               of the ratio p/R, i.e. of the distance p
(LV switchboard, duct...). We thus need        asymmetrical).                                        between the supports. p must be chosen
to reason on the «busbar structure».           The complete study was conducted by                   so that the ratio is outside the hatched
There are two standard methods for             Baltensperger and leads to an                         zone for the accepted factor Vσ.
securing busbars: flush mounting and
simple support. However in reality the
insulating elements support the
conductors, which results in a
combination of these two methods               Vσ                                                           - Vσ: stress factor
(see fig. 11).
The large number of parameters to be                                                                        - p: distance between supports
                                                5
considered makes a complete study of                                                                                  EJ
                                                                                                            - R=4          103
these phenomena complex.                                                                                             M ω2
The starting point for such a study is          4
the general equation applied to a
conductor assumed to have an elastic
behaviour:                                      3

                                                                                        resonance zone
                                                2


     a                b               c
                                                1


                                                                                                                                             p
                                                                 1         2.24 2.45    3     3.55   4             5.22      6.12     7      R

fig. 11: the various busbar securing
methods: by flush mounting (a), by simple
support (b) and a combination of both (c).     fig. 12: stress factor Vσ to be anticipated as a function of the ratio p/R.




                                                                                                     Cahier Technique Merlin Gerin n° 162 / p.11
4. application to LV three-phase busbars


In this chapter the authors define how            In practice, sizing requires determination   The details of each stage are described
the above theoretical considerations              of the distance between the supports         below for a busbar consisting of several
are taken into account for two LV items           and thus the number required, for a          rectangular cross-section bars per
of equipment, namely LV switchboards              specific busbar and support technology.      phase.
and prefabricated electrical ducts of the
Canalis and Victa Dis type.                       Practical calculation procedure              I - Basic data for carrying out the
                                                  The method to be followed is                 calculation
                                                  summarised in the chart below:               c dimension and shape of a conductor
case of busbars in LV                                                                          (for example for a bar, its thickness a
switchboards                                               I. Definition of basic data
                                                                                               and its width b in m.)
The three-phase busbar of a LV electric                                                        c number of conductors per phase: n.
switchboard is made up of a set of                          II. Calculation of forces
conductors grouped by phase and held                                                           c root mean square value of the short-
in place by supports.                                                                          circuit current: Isc in kA.
                                                   III. Calculation of the distance between    c type of fault: two or three-phase.
It is characterised by:
                                                      supports based on stresses on the
c the shape of the conductors,                                                                 c distance between phase centres: dph
                                                       conductor with the greatest stress
c the relative layout of the phases,                                                           in m.
c the arrangement of the conductors in
                                                                                               c conductor securing method in the
the same phase,                                        IV. Calculation of the distance
                                                                                               supports (flush mounting or simple
c the type of support and the conductor             between supports based on stresses
                                                                                               support).
securing method (insulating bars,                               on supports.
                                                                                               This data is taken into account by a
combs, insulating rods...).
                                                                                               coefficient ß:
The various elements making up the                   V. Determination of the maximum           ß = ß1 for all the conductors of a
busbar system must be sized to                                                                 phase,
                                                      distance between supports, and
withstand the electrodynamic forces
                                                    verification of the vibration behaviour    ß = ß2 for a conductor belonging to
which appear when a short-circuit
                                                                 of the busbar.                one phase,
occurs (see fig. 13).
                                                                                               c elastic limit of the conductor:
                                                                                               Rp0.2 in N/m2
                                                                                               (Rp0.2 = 125 x 106 N/m2 for 1050 type
                                                                                               aluminium and Rp0.2 = 250 x 106 N/m2
                                                                                               for copper).
                                                                                               c characteristics of supports:
                                                                                               mechanical withstand Rm (in N/m2)
                                                                                               according to the type of stress, and
                                                                                               cross-section of the stressed support
                                                                                               Sm (in m2).
                                                                                               II - Calculation of forces
                                                                                               Each conductor of a phase is subjected
                                                                                               to a force due to the actions between
                                                                                               phases and to the actions of the other
                                                                                               conductors of the same phase. The
                                                                                               maximum force is exerted on the most
                                                                                               external conductors of the central phase.
                                                                                               This conductor is subjected:
                                                                                               c firstly to the force resulting from the
                                                                                               other two phases:
                                                                                               F1/ l
                                                                                               = 0.87 (or 1) 2 10-7 k1 (2.2 Isc)2 1/dph)
fig. 13: busbar of a Masterbloc LV switchboard, designed to withstand the effects of a 80 kA
                                                                                               0.87 : if the fault is three phase
short-circuit current (Merlin Gerin).
                                                                                               1 : if the fault is two phase




Cahier Technique Merlin Gerin n° 162 / p.12
(force per unit of length of the busbar in
N/m).
k1 = Dwight's coefficient allowing for the
shape of all the conductors of the                                 phase 1                 phase 2                   phase 3
phase.
                                                                                            d'            d ph
This coefficient, parameterised by the
ratios height (h)/width of a phase (I')
and dph/width of a phase, can be                                                       b                         h
calculated or read on charts.                             d1 → 2
Isc = root mean square value of the                       d1 → 3                      a                                 l'
short-circuit current in kA.
dph = distance between phase centres          here b = h
in m.
                                              k1 = f(h, I', dph)
The multiplying factor 2.2 is used to
                                              k2 = f(a, b, d')
calculate the peak value of the short-
circuit current.
c secondly to the force of attraction         fig. 14: parameters considered to establish the equation for the force of attraction between
(current in the same direction) resulting     busbar conductors
from the other conductors of the phase
considered (see fig. 14), if these are
mechanically linked:                          configuration; bars of the same phase              a distance d equal to the smallest value
                                              are flush mounted and the three                    of d1 and d2:
F2 / l = ∑ F21→i / l (in N / m)               phases positioned (see fig. 15).                   d i min (d1,d2).
           i                                  «Busbar deformation» criterion                     Moreover you must ensure that this
Equation of the same form as the one          The bar with the greatest stress must              distance does not generate resonance
above, but taking into account the            not be deformed. However a slight                  phenomena.
following three parameters:                   residual deformation is accepted                   This calculation procedure complies
d1 → i = centre distance from                 according to a coefficient q defined by            with the recommendations of the
conductor 1 to conductor i in m,              the IEC 865 standard.                              IEC 865 standard (1986) dealing with
n = number of conductors per phase,           The above formula includes d1. This                calculation of the effects of short-circuit
k2 = Dwight's coefficient for the phase       distance between supports can be                   currents as regards both the thermal
conductor.                                    determined from a maximum stress                   and mechanical aspects.
                                              level at the conductors which must not             Although these calculations do not
III - Calculation of the distance             be exceeded, such that σ = q.Rp0.2 (for            replace real-life tests, they are vital for
between supports based on stresses            example q = 1.5).                                  designing new products and for
on the conductor with the greatest                                                               satisfying specific cases.
                                              IV - Calculation of the distance
stress
                                              between supports based on stresses
The conductor with the greatest stress        on supports
must withstand the stress:                    The supports must therefore withstand




                                                                                           ,,,,,
                                                                                             ,,
                                                                                           ,,,,,
σ = σ1+ σ 2                                   the stresses linked to the force F1.
       β1 (F1/ l) d12       β2 (F2 / l) d12                                                                                       flush




                                                                                             ,,
                                                                                           ,,,,,
                                              «Support break» criterion:                                                          mounted
   =                    +                                                                                                         conductors
               8Z                8 Z0              Rm Sm
                                              d2 =
F1/ l and F2/ l = forces in N/m,
                                                   α F1/ l
                                              where
d1 = distance between two supports
                                              α = constant whose value depends on
in m,
                                              the securing method and the number of
Z0 = resistance module of a bar in m3,        supports.                                                            phases in
Z = resistance module of a phase                                                                                 simple support
                                              V - Determination of the maximum
in m3,
                                              distance between supports, and
ß1 = 0.73 (simple support coefficient),       verification of the busbar vibration
ß2 = 0.5 (flush mounting coefficient).        behaviour                                          fig. 15: configuration of a busbar for
These values are given by way of                                                                 coefficients ß1 = 0.73 (simple support) and
                                              In order to withstand electrodynamic
                                                                                                 ß2 = 0.5 (flush mounting).
guidance for a specific busbar                forces, the supports must be placed at




                                                                                                 Cahier Technique Merlin Gerin n° 162 / p.13
Calculation example
I. definition of basic data
c conductors
flat copper bars
thickness a = 5 mm
width b = 100 mm
securing: flush mounted bars
c each phase is made up of
n = 3 bars, with a 5 mm spacing (d' = 10 mm)
c distance between phase centres
dph = 95 mm
c three-phase fault Isc = 80 kA rms
c elastic limit of copper
Rp0.2 = 250 x 106 N/m2
c mechanical withstand of support
Rm = 100 x 106 N/m2
c cross-section of support subjected to tensile stress
Sm = 150 x 10-6 m2
II. calculation of forces
c between phases
F1/ l = 0.87 2 10-7 k1 (2.2 Isc) 2 1/dph                                F1/ l =
k1 : Dwight's coefficient, function of the ratios b/(2n - 1) a          0.87 x 2 x 10-7x 0.873 x (2.2 x 80 x 103) 2 x 1/95 x 10-3
and dph/(2n - 1) a                                                      F1/ l = 49 530 N/m = 4 953 daN/m
k1 (100/5 5, 95/5 5) = 0.873
c between bars of the same phase
particularly on the external bars of the central phase
F2 / l = ∑ F21,i / l                                                    F2/ l =
         i                                                              2 x 10-7x (2.2 x 80 x 103/n) 2 x [0.248/10 x 10-3 + 0.419/20 x 10-3]
1 index of the first bar                                                F2/ l = 31 490 N/m = 3 149 daN/m
i = 2 and 3 index of the two other bars of the phase
F2 1, i / l = 2 10-7 k21,i ( 2.2 Icc/n ) 2 1/d
 d1→i : distance between the axis bars 1 and i
k2 1, i: Dwight's coefficient as a function of ratios b/a and d1→i /a
k2 1, 2( 100/5, 10/5 ) = 0.248
k2 1, 3( 100/5, 20/5 ) = 0.419
III. Calculation of the distance between supports based on
stresses on the conductor with the greatest stress (elastic
limit of conductor)
σ = β1 (F1/ l ) d12/8 Z + β2 (F2/ l ) d12/8 Z0
σ = 1.5 Rp0.2
d12 = 1.5 Rp0.2 / [β1 (F1/ l )/8 Z + β2 (F2/ l )/8 Z0 ]                 d12 = 1.5 x 250 x 106 / [ 0.5 x (49 530)/8 x 1.25 x 10-6 +
β1 = β2 = 0.5                                                           0.5 x (31 490)/8 x 4.2 x 10-7]
Z0 = b a2/6 = 4.2 10-7 m3
Z = n Z0 = 3 Z0 = 1.25 10-6 m3                                          d1 = 0.229 m = 229 mm
IV. Calculation of the distance between supports based on
stresses on the supports (elastic limit of supports)
d2 = Rm Sm/(F1/ l )                                                     d2 = 100 x 106 x 150 x 10-6/(0.5 x 49 530)
α = 0,5                                                                 d2 = 0.604 m = 604 mm
V. determination of the maximum distance between supports
d = minimum entre d1 et d2                                              d < 229 mm




Cahier Technique Merlin Gerin n° 162 / p.14
Standards and tests                                be qualified by calculation from               not the busbar impedance is taken into
There are two test categories for LV               an T.T.A. structure.                           account during calculation. In practice:
equipment, namely:                                 With respect to short-circuit current          c calculation performed with a voltage
                                                   withstand, an extrapolation method for         equal to the operational voltage, at the
c development tests assisting with
                                                   the P.T.T.A. has been defined by the           entrance to the
design,
                                                   technical report IEC 1117 (1992).              switchboard ⇒ presumed value of Isc;
c certification tests.                                                                            c calculation performed, at extra-low
                                                   Complete certification in short-circuit
The latter are part of a set of tests              current withstand requires three tests:        voltage, at the end of the busbar at the
known as «type tests» whose reports                c a three-phase short-circuit current          short-circuit point ⇒ real value.
are frequently demanded for a product              withstand test;                                It is obvious that for the same
defined as a «Type tested assembly»                c a withstand test for a short-circuit         announced value of the short-circuit
(T.T.A.).                                          current between the neutral and the            current strength, the second case is far
This designation, which requires tests,            nearest phase. Note that if the neutral        more restrictive. The difference may
thus forms an additional guarantee for             has the same cross-section as the              range from 20 to 30% according to the
users. However, despite this constraint,           other phases and if the distance               circuit.
manufacturers develop products which               between the neutral and the nearest            For the phase-neutral test, the value of
allow them to valorise their know-how.             phase is the same as the distance              the short-circuit current corresponds to
The type tests defined by the standards            between phases, this test corresponds          60% of the value of the current
IEC 439-1 (1992) and 2 (1987) or                   to a two-phase short-circuit;                  (prospective or real) of the three-phase
NF 63-421 (1991) total 7 (439-1) and               c a withstand test for a short-circuit         test.
10 (439-2) respectively.                           between a phase and the protective             Many manufacturers (including
As regards short-circuit withstand,                conductor.                                     Merlin Gerin and Telemecanique)
which is the subject of this document,             For each test the manufacturer must            currently tend to perform these tests in
these standards specify both the test              specify the root mean square value of          real current. Moreover, to ensure that
conditions to be complied with and the             the short-circuit current and its duration,    these tests are representative of the
standardised value of the coefficient              normally 1 s (to verify the thermal            most unfavourable tests possible during
connecting the peak value to the root              constraint linked to the short-circuit         a short-circuit, the following points must
mean square value of the short-circuit             current).                                      be complied with:
current (see fig. 16).                             As regards the value of the short-circuit      c presence of an asymmetrical state at
                                                   current for the three-phase test, two          least on one of the three phases;
If the system considered varies only
                                                   values must be identified: the                 c presence of at least one joint or
slightly from the reference system
                                                   prospective value and the real value.          fishplate on the tested busbar;
(T.T.A.), it is known as a «Partially type
                                                                                                  c creation of a bolted short-circuit;
tested assembly» (P.T.T.A.) and it can             Their difference is due to whether or
                                                                                                  c consideration of vibrating
                                                                                                  phenomena, while maintaining the fault
                                                                                                  for at least ten cycles, i.e. 200 ms at
                                                                                                  50 Hz; this time is often extended to 1 s
                                                                                                  to check thermal withstand at the same
root mean square value of the                      cos ϕ                  n                       time (IEC 439-1).
short-ciruit current (kA)
                                                                                                  The various test stages are:
Ii 5                                               0.7                   1.5                      c calibration circuit by short-circuiting
5 < I i 10                                         0.5                   1.7                      the transformer outputs;
10 < I i 20                                        0.3                   2
                                                                                                  c connecting the busbar to the platform
20 < I i 50                                        0.25                  2.1
                                                                                                  transformer;
50 < I                                             0.2                   2.2
                                                                                                  c setting up the short-circuit (specific
                                                                                                  part connecting all the bars) on the
                                                                                                  busbar;
fig. 16 : standardised value of the coefficient n connecting the peak value to the root mean      c short test (roughly 10 ms) to
square value of the short-circuit current; n correspond to coefficient 2 κ defined in chapter 3
                                                                                                  determine busbar impedance;
(ie. IEC 439-1).
                                                                                                  s 1 s withstand test on the assembly.




                                                                                                  Cahier Technique Merlin Gerin n° 162 / p.15
case of prefabricated ducts                        The electrodynamic forces developing        conductors. Some ducts sandwich up
                                                   when short-circuits occur, observe the      to five conductors per phase.
of the Canalis or Victa Dis                        laws stated above and result in the         Designers can then either:
type                                               deflection of conductors between            v leave the conductors of the same
Construction of prefabricated three-               insulators and in an overall vibration.     phase grouped together,
phase busbars of the Canalis or                    The shape and cross-section of the          v or insert the elementary phase
Victa Dis type, designed for current               conductors result from the best             conductors in an orderly manner (1-2-3)
transmission and distribution (see                 possible balance achieved between:          + (1-2-3) + (1-2-3) to obtain the
fig. 17) complies with proper                      v temperature rise of conductors;           «sandwiched» configuration (see
procedures and with specific standards,            v acceptable voltage drop;                  fig. 18).
the main ones of which are                         v production cost.
                                                                                               This type of design is ideal for
the IEC 439-1 and 2 (international)                With the following vital requirements for   horizontal current distribution.
and UL 857 (United States).                        mechanical withstand: that conductor
                                                   deflection continues to be elastic (no      c flattened design (1000 to 5000 A):
Design                                                                                         In this layout, the rectangular cross-
                                                   permanent deformation after a short-
The techniques implemented vary                                                                section conductors coated with an
                                                   circuit) and does not abnormally reduce
according to the current ranges                                                                insulating sheath are kept in contact
                                                   the insulation level (between phases or
considered, especially for large                                                               all along the duct, just as in a cable
                                                   between phases and earth) during the
currents exceeding 100 A.                                                                      (see fig. 18). Conductors are clamped
                                                   short-circuit transient period in which
There are currently three main duct                the electrodynamic phenomena are            to ensure the necessary heat
designs:                                           created.                                    exchanges.
v standard,                                        In practice this is obtained by adjusting   To simplify manufacture, conductors
v sandwich,                                        the distance between insulators.            normally have constant thickness, and
v flattened.                                                                                   only their width varies according to
                                                   c sandwiched design (1000 to 5000 A):
c standard design (100 to 800 A)                   Beyond a certain current, 1000 A, in        nominal busbar current strength (up to
The conductors are placed in a metal               order to remain within acceptable heat      roughly 250 mm). For high currents,
envelope and maintained at regular                 exchange conditions and dimensions          two or even three conductors per
intervals by comb shaped insulators                for the duct, the current of the same       phase, but not sandwiched, are
(see fig. 18).                                     phase is distributed over several           required.




                                                                                               a)
                                                                                                                                           N
                                                                                                                                           L3
                                                                                                                                           L2
                                                                                                                                           L1




                                                                                               b)



                                                                                                                                           N
                                                                                                                                           L3
                                                                                                                                           L2
                                                                                                                                           L1



                                                                                               c)                                          N
                                                                                                                                           L3
                                                                                                                                           L2
                                                                                                                                           L1



                                                                                               fig. 18: the various prefabricated three-
                                                                                               phase busbar designs: standard (a),
fig. 17: Canalis (Telemecanique) 3000 A electrical distribution prefabricated busbar.          sandwiched (b) and flattened (c).




Cahier Technique Merlin Gerin n° 162 / p.16
The electrodynamic forces (distributed               various forces to which conductors are              The calculations shown in the box (see
loads) when a short-circuit occurs are               subjected.                                          fig. 19) evaluate the mechanical stress
balanced in these busbars by the                     The structure studied has the following             of the elementary conductors
reaction of the envelope sheet metal.                characteristics:                                    (according to current direction) of
Its thermal behaviour means that this                c In = 3000 A,                                      phases 1 and 2 for a phase 1/phase 2
type of design is ideally suited to                  c three conductors/phase, i.e.                      two-phase short-circuit with correction
transmission of horizontal or vertical               1000 A/conductor,                                   of the geometric incidence in
current.                                             c conductor cross-section =                         accordance with Dwight's chart.
Distribution of electrodynamic                       90 mm x 6 mm,                                       Partial conclusions:
forces                                               c material = aluminium or copper,                   With the standard layout, an increase
This paragraph uses a simple, concrete               c distance between conductor                        and large dispersion of the forces
example to visualise and quantify the                centres = 18 mm.                                    applied to the various conductor




    1   1'   1"      2    2'   2"     3    3'   3"
                                                     standard design


    + + +            -    -    -                d

                                                                    (         )
                                                               F = f i2 ,cosϕ with i = I / 3


                                                                               1                                                   
                                                                ∑Fph1= F  d
                                                                          
                                                                         
                                                                                          1
                                                                                    0.42 + 
                                                                                         2 d
                                                                                                     1
                                                                                               0.62 − 
                                                                                                    4 d
                                                                                                                1
                                                                                                          0.83 − 
                                                                                                               5 d
                                                                                                                           1
                                                                                                                     0.87 − 
                                                                                                                          6 d
                                                                                                                                0.92 
                                                                                                                                    

                                                                            F
                                                                        =     0.19 ⇒ K1= 0.19
                                                                            d

                                                                                                                                     
                                                                ∑Fph1' = F  − d            1       1        1        1
                                                                                    1
                                                                                        0.42 +   0.42 −     0.75 −     0.83 −     0.87 
                                                                                           d      3 d      4 d      5 d     

                                                                              F
                                                                        =−      0.63 ⇒ K1'= 0.63
                                                                              d

                                                                                                                                       
                                                                ∑Fph1" = F  − d             1        1        1        1
                                                                                    1
                                                                                        0.42 +     0.62 −     0.62 −     0.75 −     0.83 
                                                                                           2 d      2 d      3 d      4 d     

                                                                              F
                                                                        =−      1.49 ⇒ K1"= 1.49
                                                                              d

    1   2    3       1'   2'   3"     1"   2"   3"
                                                     sandwich design

    +    -           +    -           +    -
                                                                                                                   1   1 
                                                                ∑Fph1= F − d                1        1
                                                                                    1
                                                                                        0.42 +     0.83 −     0.87 −      1 −     1
                                                                                           4 d      5 d       8 d   9 d 

                                                                             F
                                                                        =−     0.37 ⇒ K1= 0.37
                                                                             d

                                                                                                                              1 
                                                                ∑Fph2 = F + d               1        1        1
                                                                                    1
                                                                                        0.42 −     0.75 +     0.83 −     0.97 +      1
                                                                                           3 d      4 d      7 d       8 d 
                                                                              F
                                                                         =+     0.36 ⇒ K2 = 0.36
                                                                              d



fig. 19: mechanical stress of phase 1 and 2 elementary conductors.




                                                                                                         Cahier Technique Merlin Gerin n° 162 / p.17
elements are observed, whereas for the          Application of this technology has the            This technology is particularly used in
sandwich layout, forces remain more or          following practical limits:                       the following current ranges:
less the same for each conductor                v 1250 A, in branch-off,                          v 16 to 400 A in branch-off,
element.                                        v up to 6000 A in splicing.                       v 40 to 1000 A in splicing.
In this example, the difference in              NB :                                              c electrodynamic force withstand
mechanical stress has a ratio of 1 to 5         Some articulated bends produced in                Whereas «bolted» technology
in favour of the sandwich layout.               the same plane use the «bolted»                   imposes on the elements of the
Moreover, this layout offers another            technology principle.                             structure in question the same
advantage as for as voltage drop is             c «contact» technology                            electrodynamic forces as for the
concerned: «sandwiching» of phases              Current conducted using parallel-                 busbars in LV switchboards,
causes a reduction in the magnetic              connected contact fingers.                        «contact» technology benefits from
induction resultant and thus in                                                                   these forces.
                                                As a first approach, the current is
reactance,... i..e. in voltage drop.            distributed in proportion to the number           The layout normally chosen for the
Branch-offs and splicing                        of parallel contacts. Each contact point          contact fingers or «pawls» is
Two technologies are normally chosen            has a static force F (developed by an             illustrated in the drawing in figure 20
to sample current at the branch-offs or         external spring) whose sizing results             on which it is clearly shown that the
to conduct it in the splice bars of a           from a compromise between the level               currents flowing in opposing pawls
prefabricated transmission and                  of the required contact resistance to             run in the same direction. The
distribution line. These are «bolted»           ensure nominal current flow without               electrodynamic forces (distributed
technology and «contact» technology.            abnormal temperature rise, and the                loads) developed along the pawls
                                                friction force withstand during                   and calculated using the above
c «bolted» technology                           conductor expansion.
The connections are made from special                                                             methods thus tend towards an
                                                With this in mind, we should note the             attraction. They consequently
bolted pads provided in the equipment           advantage of lubricating the elastic              reinforce the contact force and
design stage.                                   contacts or of using, for mounted                 oppose the repulsion force of
The above laws are also applied for             contacts, silver/graphite type                    contacts which has as its origin
sizing the pads and insulators.                 combinations.                                     striction of the current lines in the
                                                                                                  vicinity of the contact point
                                                                                                  (see fig. 21). This is the self-
                                                                                                  compensation principle (see fig. 20).
cross-section BB                                k is the shape factor, to be read off the chart
                      l                         in figure 2, for a conductor with a global
                                                                                                  Electrodynamic force tests
                       A                        cross-section a x b.                              Type tests, specific to ducts, are
                                                If we write Fa Fr so that compensation is         defined by the IEC 439-2 and
     I                                          achieved, the result is:                          NF C 63-411 standards.
                                            d                     2                               The main difference compared with
                                                    u        + 1 − 1
                                                 l          3
                                                         n k                                    «LV switchboards» lies in the short-
                                                 d
                                                                                                  circuit test conditions which specify
                       A                        For example, for k = 0.8, the ratio I/d must
                                                                                                  that the tests must be performed on an
                                                reach:
                                                                                                  installed line no more than six metres
cross-section AA           B                    4.6      for 1 contact jaw          (n = 1)
                                                                                                  long with at least one splice joint and a
                                                2.7      for 2 contact jaws         (n = 2)
                                                                                                  bend (see sketch in fig. 22).
                                                1.4      for 5 contact jaws         (n = 5)
                                                0.95     for 10 contact jaws        (n = 10).
                                                Although it may seem interesting to increase
                           B                    the number of parallel contact jaws n, we are
                                                                                                                          I
                                                quickly limited by technological
In the calculation, we consider the forces on   considerations as well as by differences in
each half-contact jaws, of cross-section        resistance and reactance between adjacent                R                              2r
a x b, grouping all the contact elements of     contact jaws which do not allow even current
jaws.                                           distribution between each other such as is
If n is the number of parallel contact jaws,    assumed by the calculated value of I/d.           A repulsion force F is exerted between the
the total repulsion force on the half-contact   We must therefore take a safety margin on         two conductors:
jaws is:                                        the calculated value of I/d, as large as the
Fr = 2 n 3 10-7 (I2n)2
                                                number of parallel contact jaws is high. In
The force of attraction which has to                                                              (In = natural log and r = contact point radius
                                                practice, there are applications of up to
compensate it is:
               I  k l  1+ d − d 
                   2               2            2 x 12 parallel contacts which can withstand      calculated with R. Holm's formula).
 Fa = 2 10 −7                                 acceptable short-term currents of the order
               2      d       l 2
                                      l        of 50 kA RMS - 1 s.                               fig. 21: example of striction of current lines in
                                                                                                  the vicinity of the contact point between two
fig. 20: the self-compensation principle.                                                         cylindrical current-carrying elements




Cahier Technique Merlin Gerin n° 162 / p.18
                                                     test source




                                                     source connecting device




                                                     sheathed busbar length




              maximum                                splice joint (at least one)
              imposed
              length
              =6m




                                                     sheathed bend (at least one)


                                                     short-circuit device




fig. 22: sketch showing a prefabricated busbar line such as defined by the standards for the
type tests.




                                                                                               Cahier Technique Merlin Gerin n° 162 / p.19
5. conclusion


The high electrodynamic forces                It is thus advantageous for installers     manufacturers can support in view of
occurring on a short-circuit and the          and/or users to choose equipment           the necessary infrastructure and costs
material damage that they can cause           presenting a maximum guarantee             involved.
justify the importance attached to            (T.T.A.) or made up of modified            Design modifications from the type
mechanical withstand of busbars. An           standard elements, mounted in the          tested cases are, however, possible.It
importance all the more vital as busbar       factory and tested (P.T.T.A.).             is in this respect, to a certain extent,
withstand failure requires at the very        In both cases, the importance of testing   that the calculation approach and the
least replacement of these busbars and        is obvious. However such tests call for    manufacturer's knowhow can take over
thus shutdown of the installation.            considerable investment that only major    from the experimental approach.




6. bibliography


Standards                                     c Operating dependability and LV           c Calculation of three-phase busbar
c IEC 439-1: Low-voltage switchgear           electric switchboards                      installations in view of withstand of
and controlgear assemblies. Type-             Cahier Technique n° 156 - O. BOUJU         electrodynamic forces
tested and partially type-tested              c Calculation of short-circuit currents    R. MASCARIN
assemblies.                                   Cahier Technique n° 158                    Revue générale de l'Electricité
c IEC 439-2: Low-voltage switchgear           R. CALVAS, B. DE METZ-NOBLAT, A.           RGE, August 1957.
and controlgear assemblies. Particular        DUCLUZAUX, and G. THOMASSET                c On the establishment of formulae
requirements for busbar trunking                                                         designed to determine force per unit
systems (busways). (NF C 63-411).             Various publications                       length in case of a short-circuit affecting
c IEC 865: Short-cicuit currents-             c Elektodynamische Beanspruchung           a three-phase busbar in a ribbon
calculation of effects.                       von parallelen Leitern,                    arrangement.
                                              or                                         R. MASCARIN.
c IEC 909: Short-circuit current              Electrodynamic effects on parallel lines   Revue générale de l'Electricité
calculation in three-phase a.c. systems.
                                              P. BALTENSPERGER                           RGE, March 1959.
c IEC 909-1: Short-circuit current            Bulletin Schweiz Elektotechn Verein        c Uber den Einflu von
calculation in three-phase a.c. systems.      n° 25, 1944                                Resonanzerscheingungen auf die
Factors for the calculation of short-
                                              c Sectioned busbars in first and second    mechanische Kurzschlu festigkeit von
circuit currents in three-phase a.c.          category installations                     biegesteifen Stromleitern.
systems according to IEC 909.
                                              R. ROLS.                                   P. SIEBER
c IEC 1117: Method for assessing the          Four-part paper published in the Revue     AEG Mitteilungen n° 49, 1959
short-circuit withstand strength of           de l'Aluminium n° 212 - 213 - 214 - 215,   c Mechanical forces on current-carrying
partially type-tested assemblies (PTTA).      1954                                       conductors.
Merlin Gerin Cahiers Techniques               c Transmission of high currents in LV      E.D. CHARLES
                                              and MV ac current - 2nd part -             Proceedings IEE, vol. 110, n° 9,
c Thermal study of LV electric
                                              P. BEIGBEDER                               September 1963
switchboards                                  Bulletin Etudes et Réalisations n° 43,
Cahier Technique n° 145                                                                  c Electrodynamic forces appearing in
                                              1957                                       electric substations on a short-circuit
C. KILINDJIAN
                                                                                         G. SCHAFFER
                                                                                         Revue Brown Boveri, 1970.




                                                                                         Réal.: Sodipe - Valence - Photo.: IPV - Grenoble
                                                                                         Edition: DTE - Grenoble
Cahier Technique Merlin Gerin n° 162 / p.20                                              10-96 - 2500 - Printing.: Clerc
                                                                                         Printed in France